Fourier Series and Half Sine Series

Question

I'm having a doubt on solving the below question.

Question: drive google(dot)com/file/d/1OOL_WecZgLk3fcd65UxVzec4T9Y7ztuB/view?usp=sharing

Apparently we can write t as, t = (4/pi)* ( (-1)^(n+1) ) * [sigma n=1-infinity { (sin((n*pi*t)/2) } ]

Hw can I find the Fourier series of f(t) = t^2?

Can I take b_n = (4/pi)*((-1)^(n+1)) and use Parsvel's identity to find the Fourier series of f(t) = t^2, which is the square of t?

Or else, if we're to use the property that t^2 = integration of 0.5*t, how can we do it?

What's the most ideal way to solve this problem?

Thank you in advance for your help!

Answer 1

To find the Fourier sine series of f(t) = t² on (0,1), you cannot square the sine series for t. Squaring a Fourier series does NOT produce the series of the squared function — it produces a convolution of coefficients. So bₙ ≠ (coefficient of t)², and Parseval’s identity cannot be used to derive the series either.

The correct approach is to compute the sine coefficients directly:

bₙ = 2 ∫₀¹ t² sin(nπt) dt

Evaluating this (using integration by parts twice) gives:

bₙ = -[2(-1)ⁿ]/(nπ) – [4(1 – (-1)ⁿ)]/(n³π³)

So the Fourier sine series of t² is:

t² = Σ (from n=1 to ∞) [ -2(-1)ⁿ/(nπ) – 4(1 – (-1)ⁿ)/(n³π³ ) ] sin(nπt)

This is the clean and standard way to find the expansion. You cannot obtain the series for t² by squaring the series for t, but you can check your result afterward using Parseval if needed.